6-5: Conditions of Special Parallelograms Page 412 10) 29 11) 25 12) 13) 14.5 14) 31.5 15) 𝒎∠𝑽𝑾𝑿=𝟏𝟑𝟐° 𝒎∠𝑾𝒀𝑿=𝟔𝟔° 18) 𝒎∠𝟏=𝟐𝟗° 𝒎∠𝟐=𝟔𝟏° 𝒎∠𝟑=90° 𝒎∠𝟒=29° 𝒎∠𝟓=𝟗𝟎° 19) 𝒎∠𝟏=𝟓𝟒° 𝒎∠𝟐=36° 𝒎∠𝟑=54° 𝒎∠𝟒=108° 𝒎∠𝟓=72° 20) 𝒎∠𝟏=𝟗𝟎° 𝒎∠𝟐=45° 𝒎∠𝟑=45° 𝒎∠𝟒=45° 𝒎∠𝟓=45° 21) 𝒎∠𝟏=𝟏𝟐𝟔° 𝒎∠𝟐=27° 𝒎∠𝟑=27° 𝒎∠𝟒=126° 𝒎∠𝟓=27° 22) 𝒎∠𝟏=𝟓𝟓° 𝒎∠𝟐=55° 𝒎∠𝟑=55° 𝒎∠𝟒=70° 𝒎∠𝟓=55° 23) 𝒎∠𝟏=64° 𝒎∠𝟐=64° 𝒎∠𝟑=26° 𝒎∠𝟒=90° 𝒎∠𝟓=64° 24) Always 25) Sometimes 26) 27) 28) 29) 30) 31) 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

Properties of Kites and Trapezoids Section 6-5 Geometry PreAP, Revised ©2013 viet.dang@humble.k12.tx.us 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Definitions Kite is a quadrilateral with exactly two pairs of congruent consecutive sides. Trapezoid is a quadrilateral with exactly one pair of parallel sides Each parallel side is called a base Base angles of a trapezoid are two consecutive angles whose common side is a base 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Properties of Kites A kite is a quadrilateral that has two pairs of consecutive congruent sides, but opposite sides are not congruent. Just as in an isosceles triangle, the angles between each pair of congruent sides are vertex angles. The other pair of angles are nonvertex angles. If a quadrilateral is a kite, then its diagonals are perpendicular If a quadrilateral is a kite, then exactly one pair of opposite angles are congruent 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Kites If a quadrilateral is a kite, then the nonvertex angles are congruent. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Kites If a quadrilateral is a kite, then the diagonal connecting the vertex angles is the perpendicular bisector of the other diagonal. and CE  AE. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Kites If a quadrilateral is a kite, then a diagonal bisects the opposite non-congruent vertex angles. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Example 1 In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mBCD. mBCD + mCBF + mCDF = 180° mBCD + 52° + 52° = 180° mBCD = 76° 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Example 2 In kite ABCD, mDAB = 54°, and mCDF = 52°. Find mABC. ADC  ABC mADC = mABC mABC + mBCD + mADC + mDAB = 360° mABC + mBCD + mABC + mDAB = 360° mABC + 76° + mABC + 54° = 360° 2mABC = 230° 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Your Turn In kite PQRS, mPQR = 78°, and mTRS = 59°. Find mQRT. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

Properties of Trapezoids The parallel sides are called bases The non-parallel sides are called legs A trapezoid has two pairs of base angles 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Trapezoids A. If a quadrilateral is a trapezoid, then the consecutive angles between the bases are supplementary. If ABCD is a trapezoid, then x + y = 180° and r + t = 180°. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

Midsegment Trapezoids B. A midsegment of a trapezoid is a segment that connects the midpoints of the legs of a trapezoids. If ABCD is a trapezoid, then x + y = 180° and r + t = 180°. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Trapezoids C. If a trapezoid is isosceles, then each pair of base angles is congruent. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Trapezoids D. The midsegment of a trapezoid is parallel to each base and its length is one half the sum of the lengths of the bases. If 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Example 3 KP = 21.9m and FM = 32.7m. Solve for FB. KP = FM KJ = 32.7 KB + BP = KB BP = 10.8 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Example 4 Find the value of a so that PQRS is isosceles. S  P mS = mP 2a2 – 54 = a2 + 27 a2 = 81 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Your Turn Find the value of x so that PQST is isosceles. 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Example 5 Find EH. 1 16.5 = (25 + EH) 2 33 = 25 + EH 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Your Turn Solve for LP 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms

6-5: Conditions of Special Parallelograms Assignment Pg 432: 14-25, 27-36 all (omit 34) 12/3/2018 10:05 PM 6-5: Conditions of Special Parallelograms